= Solution
Under <Fermi-surface renormalization-group scaling>, tangential momentum $k$ is fixed while
$$
\omega\mapsto s\omega,
\qquad
\ell\mapsto s\ell,
\qquad s\to0.
$$
In frequency space the quadratic action is invariant when
$$
\psi(\omega,k,\ell)\mapsto s^{-3/2}\psi(\omega,k,\ell);
$$
equivalently, in the time representation $t\mapsto s^{-1}t$ and $\psi(t,k,\ell)\mapsto s^{-1/2}\psi(t,k,\ell)$. The free <Fermi liquid> action is therefore marginal.
The deformation $\mu(k)\psi^\dagger\psi$ has no factor of frequency or normal momentum, so it is relevant. It shifts the zero of the quasiparticle energy from $\ell=0$ to $\ell=\mu(k)/v_F(k)$ and hence deforms the <Fermi surface>.
A generic quartic interaction is irrelevant because momentum conservation fixes a normal component and leaves an additional positive power of $s$. It can be marginal only when all four momenta remain on the Fermi surface while satisfying
$$
\mathbf k_1+\mathbf k_2=\mathbf k_3+\mathbf k_4.
$$
For a smooth generic Fermi surface, the robust possibilities are forward or exchange scattering and the opposite-momentum BCS channel $\mathbf k_2=-\mathbf k_1$, $\mathbf k_4=-\mathbf k_3$.
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